Melody

I wasn't targeting a cheat nerdiest.

I have no more reason to think that this person is cheating than I do for most others.

I simply like to encourage people to contribute to their own solutions.

They will be more litkely to learn that way.

If they do not want to do that, well fair enough, but if they should not be help until they interact.

What have you done towards answering this for yourself?

Please show us.

\(Let \qquad ki=\frac{z_2}{z_1} \qquad k\in Z\\ \displaystyle \left| \frac{2z_1+7z_2}{2z_1-7z_2}\right|\\ =\displaystyle \left| \frac{\frac{2z_1+7z_2}{z_1}}{\frac{2z_1-7z_2}{z_1}}\right|\\ =\displaystyle \left| \frac{2+7ki}{2-7ki}\right|\\ =\displaystyle \left| \frac{(2+7ki)^2}{4+49k^2}\right|\\ =\displaystyle \left| \frac{(4-49k^2)+28ki}{4+49k^2}\right|\\ =\displaystyle \frac{\sqrt{(4-49k^2)^2+(28k)^2}}{4+49k^2}\\ =\displaystyle \frac{\sqrt{ 16+2401k^4-392k^2 +784k^2}}{4+49k^2}\\ =\displaystyle \frac{\sqrt{ 2401k^4+392k^2+16 }}{4+49k^2}\\ =\displaystyle \frac{ 49k^2+4 }{4+49k^2}\\ =1 \)

You need to check that.

Let \qquad ki=\frac{z_2}{z_1} \qquad k\in Z\\
\displaystyle \left|  \frac{2z_1+7z_2}{2z_1-7z_2}\right|\\
=\displaystyle \left|  \frac{\frac{2z_1+7z_2}{z_1}}{\frac{2z_1-7z_2}{z_1}}\right|\\
=\displaystyle \left|  \frac{2+7ki}{2-7ki}\right|\\
=\displaystyle \left|  \frac{(2+7ki)^2}{4+49k^2}\right|\\
=\displaystyle \left|  \frac{(4-49k^2)+28ki}{4+49k^2}\right|\\
=\displaystyle   \frac{\sqrt{(4-49k^2)^2+(28k)^2}}{4+49k^2}\\
=\displaystyle   \frac{\sqrt{   16+2401k^4-392k^2      +784k^2}}{4+49k^2}\\
=\displaystyle   \frac{\sqrt{   2401k^4+392k^2+16 }}{4+49k^2}\\
=\displaystyle   \frac{  49k^2+4 }{4+49k^2}\\
=1

Mostly we don't bother, but yes, moderators are able to label a post as 'best answer'

LOL Ginger,

It is easy to wear blinkers with probabilitiy questions :)

But yea, the forest was hiding behind two trees.  I like that :)

Hi Ginger,

Sorry about the late response, Ive been in flood the last couple of days and have had not electricity of internet. 

It is just luck that i saw thisat all. 

-----------------------------------

How about this combination Ginger.

Every row has a rook

and every column has a rook

So it has been counted twice.

There are 8! such combinations that have been counted twice.  

So I am stickig by my answer.  

Hi Elijah    

The parabola needs to be of the form  x^2=4ay   [The vertex is (0,0)]

a is the foacl length

\(x^2 =4ay=\frac{8}{5}y=4*\frac{2}{5}y \)  

the foacal length is 2/5 = 0.4"

So the bulb must be placed 0.4 inches from the vertex.

Hi Ginger,

I finally got a chess board out and looked at the problem.  I understand it now.

I think that the 8 rooks must be placed 1 in every column which is  8^8

or

one in every row which is 8^8

But I have to subtract all the ones that are in both every row and every column (becasue they have been added twice.)

This number is 8!

So I am now pretty certain that the answer is :

   \(\displaystyle \frac{2*8^8-8!}{64C8}= \frac{33 554 432-40320}{4426165368} = \frac{33514112}{4426165368}\approx 0.75718\%\)

Hi Chris, hot18594, Ginger and guest.    And thanks for your answers and interaction   

I think that we are all in agreement that there are 36 different possible combinations of cuts.  

9 places to cut, cut in 2 places   9C2 = 36

We are also in agreement that the sides of the triangle will be (2,4,4) or (3,3,4) units in length.

BUT where can the CUTS happen for this??

I will look at 2,4,4

1 cut 3,4,5, cut, 7,8,9,10     that would give 2,4,4

or

1,2,3,cut,5,cut,7,8,9,10      that would give 4,2,4

or

1,2,3,cut,5,6,7,cut,9,cut      that would give 4,4,2

So there are 3 ways to cut the stick to get a trinagle of sides  2,4,4

there will also be 3 ways to cut the stick to get  3,3 and 4

So there are 6 favourable ways to cut the stick.

Probability of being able to make a triangle will be    6/36   =    1/6

I think that is what Ginger was refering to.  

I assume the dice are cubes, (6 sides)

The numbers on them are 1 to 6

1+2+3+4+5+6= 21     

so prob of rolling a 1 is  1/21

so prob of rolling a 2 is  2/21

etc

To roll a 3 you have to have (1,2)  ot (2,1)

P(rolling 1 followed by 2 )= 1/21 * 2/21 = 3/441

P(rolling 2 followed by 1)= 1/21 * 2/21 = 3/441

So prob of rolling a total of 3 is   6/441  = 2/147